Applied Representation Theory : 008

Matrix Representations

We discuss representations and modules for concrete vector spaces using matrix Lie algebras and groups.

Review of Modules and their Representations

Recall that a vector space can be made into a module for either a Lie algebra or a group by the existence of a linear map known as a representation. In what follows, let $V$ be a vector space, $\mathfrak{g}$ be a Lie algebra and $G$ a group.

A Lie algebra representation is a map from said Lie algebra into the endomorphism group of the associated module, $V$:

$$\pi : \mathfrak{g} \rightarrow \mathsf{End}\, V,$$

that serves as a Lie algebra homomorphism.

A group representation is similarly a map from said group into the linear automorphisms of $V$:

$$\pi : G \rightarrow \mathsf{Aut}\, V,$$

where $\pi$ is a group homomorphism.

To understand representations, we must first understand $\mathsf{End}\, V$ and $\mathsf{Aut}\, V$. That is the goal of this lecture.

Matrix Representations

For our present purpose\(^{1}\), we shall consider vector spaces of the form $V = \mathbb{A}^{n}$, where $\mathbb{A}$ is any of $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$. In this case, we are already familiar with the relevant ingredients. 

The endormophisms are given by the so-called Hom space:

$$\mathsf{End}\,\mathbb{A}^{n} = \mathsf{Hom}_{n,n}\mathbb{A},$$

that is, the $n\times n$ matrices with components valued in $\mathbb{A}$.

In a similar vein, the group of automorphisms of $\mathbb{A}^{n}$ are given by the general linear group,

$$\mathsf{Aut}\,\mathbb{A}^{n} = \mathsf{GL}_{n}\mathbb{A},$$

which consist of matrices in $\mathsf{Hom}_{n,n}\mathbb{A}$ with nonzero determinant.

Example: Representations on $\mathbb{C}^{2}$

As a vector space, $\mathbb{C}^{2}$ can be used as a module for various groups and Lie algebras. As it turns out, there is a hierarchy to these representations. We start with the biggest - and most general representations, and specialize from there.

The Lie algebra $\mathfrak{gl}_{2}$

Per our discussion above, the endomorphisms of $\mathbb{C}^{2}$ are the $2\times 2$ matrices with complex elements,

$$\mathsf{End}\,\mathbb{C}^{2} = \mathsf{Hom}_{2,2}\, \mathbb{C}.$$

Taken with matrix multiplication, $\mathsf{Hom}_{2,2}\, \mathbb{C}$ is an associative algebra. Taking the commutator as the Lie bracket, this is precisely the Lie algebra $\mathfrak{gl}_{2}$:

$$\mathfrak{gl}_{2} = \mathsf{Hom}_{2,2}\, \mathbb{C}.$$

In other words, the most general Lie algebra that can be nontrivially represented on $\mathbb{C}^{2}$ is $\mathfrak{gl}_{2}$.

The representation of $\mathfrak{gl}_{2}$ on $\mathbb{C}^{2}$:

$$\pi : \mathfrak{gl}_{2} \rightarrow \mathsf{End}\,\mathbb{C}^{2},$$

for matrix Lie algebras is actually the identity map:

$$\pi = \mathsf{id},$$

where $\pi(g) = g$ for all $g$ in $\mathfrak{g}$.

In other words, $\mathfrak{gl}_{2}$ acts on $\mathbb{C}^{2}$ as a matrix would ordinarily act on a vector.

As we shall see below, any subalgebra of $\mathfrak{gl}_{2}$ can also be represented on $\mathbb{C}^{2}$ using essentially the same, simple representation map, $\mathsf{id}$.

Finally, it should be mentioned that $\mathbb{C}^{2}$ is not the only possible module for $\mathfrak{gl}_{2}$, far from it. But it does provide for the simplest representation. We’ll return to this issue later.

The Lie group $\mathsf{GL}_{2}\mathbb{C}$

The automorphisms of $\mathbb{C}^{2}$ are precisely those matrices in $\mathsf{Hom}_{2,2}\,\mathbb{C}$ which are invertible. This forms the familiar group, $\mathsf{GL}_{2}\mathbb{C}$:

$$\mathsf{Aut}\,\mathbb{C}^{2} = \mathsf{GL}_{2}\mathbb{C}.$$

This means that that the most general group that can be nontrivially represented on $\mathbb{C}^{2}$ is in fact the Lie group $\mathsf{GL}_{2}\mathbb{C}$.

Once again, the representation $\pi$ of $G$:

$$\pi : G \rightarrow \mathsf{Aut}\,\mathbb{C}^{2},$$

is given by the identity map:

$$\pi = \mathsf{id}.$$

So again, $\pi(g) = g$ for all $g$ in $G$.

Similarly, $\mathsf{GL}_{2}$ acts on $\mathbb{C}^{2}$ as a matrix would ordinarily act on a vector.

Perhaps not surprisingly, $\mathfrak{gl}_{2}$ is the Lie algebra associated to the Lie group $\mathsf{GL}_{2}\mathbb{C}$.

Finally, it should be mentioned that $\mathbb{C}^{2}$ is not the only possible module for $\mathsf{GL}_{2}\mathbb{C}$, far from it. But it does provide for the simplest representation. We’ll return to this issue later.

The Lie algebra $\mathfrak{sl}_{2}$

If $\mathfrak{gl}_{2}$ has complex dimension four, the Lie algebra $\mathfrak{sl}_{2}$ has complex dimension three. We’ve seen it already as the complex span of the physicists’ $\sigma$ matrices:

$$\mathfrak{sl}_{2} = \mathsf{span}_{\mathbb{C}}\left\{ \sigma_{1},\sigma_{2},\sigma_{3}\right\}.$$

$\mathfrak{sl}_{2}$ is a Lie subalgebra of $\mathfrak{gl}_{2}$, so is a subalgebra of the Endomorphisms of $\mathbb{C}^{2}$. Therefore $\mathbb{C}^{2}$ is also a module for $\mathfrak{sl}_{2}$ in the obvious way:

The representation of $\mathfrak{sl}_{2}$ is the restriction of the identity map $\mathsf{id}$ to the domain $\mathfrak{sl}_{2}$:

$$\mathsf{id}\Big|_{\mathfrak{sl}_{2}} : \mathfrak{gl}_{2} \rightarrow \mathfrak{gl}_{2},$$

or in other words

$$\mathsf{id} : \mathfrak{sl}_{2} \rightarrow \mathfrak{sl}_{2}.$$

Therefore, $\mathsf{id}$ also serves as a representation of $\mathfrak{sl}_{2}$ on $\mathbb{C}^{2}$:

As per usual, $\mathfrak{sl}_{2}$ acts on $\mathbb{C}^{2}$ as a matrix would ordinarily act on a vector.

It should be mentioned that $\mathbb{C}^{2}$ is not the only possible module for $\mathfrak{sl}_{2}\mathbb{C}$, far from it. But it does provide for the simplest representation. We’ll return to this issue later.

The Lie group $\mathsf{SL}_{2}\mathbb{C}$

Recall that the Lie group $\mathsf{SL}_{2}\mathbb{C}$ is the subset of $\mathsf{GL}_{2}\mathbb{C}$ whose matrices all have determinant one. Since the group operation is matrix multiplication, and the determinant of a product is the product of determinants:

$$\det MN = \det M \det N,$$

it is clear that $\mathsf{SL}_{2}\mathbb{C}$ is a subgroup of $\mathsf{GL}_{2}\mathbb{C}$. It is also therefore a subgroup of $\mathsf{Aut}\,\mathbb{C}^{2}$.

The the representation of $\mathsf{SL}_{2}\mathbb{C}$ on $\mathbb{C}^{2}$ is given simply by the restriction of the identity map to the subgroup $\mathsf{SL}_{2}$.

$$\mathsf{id}\Big|_{\mathsf{SL}}: \mathsf{GL}_{2}\mathbb{C} \rightarrow \mathsf{GL}_{2}\mathbb{C},$$

in other words,

$$\mathsf{id}: \mathsf{SL}_{2}\mathbb{C} \rightarrow \mathsf{SL}_{2}\mathbb{C}.$$

As per usual, $\mathsf{SL}_{2}$ acts on $\mathbb{C}^{2}$ as a matrix would ordinarily act on a vector.

As you no doubt have assessed by now, $\mathbb{C}^{2}$ can be a module for any subgroup of $\mathsf{Aut}\,\mathbb{C}^{2}$ by the same logic.

As we have learned previously, $\mathfrak{sl}_{2}$ is the Lie algebra associated to the Lie group $\mathsf{SL}_{2}\mathbb{C}$.

Finally, it should be mentioned that $\mathbb{C}^{2}$ is not the only possible module for $\mathsf{SL}_{2}\mathbb{C}$, far from it. But it does provide for the simplest representation. We’ll return to this issue later.

The Lie algebra $\mathfrak{su}_{2}$

If $\mathfrak{gl}_{2}$ has complex dimension four, and the Lie algebra $\mathfrak{sl}_{2}$ has complex dimension three, $\mathfrak{su}_{2}$ has real dimension three. We’ve seen it already as the real span of the physicists’ $\sigma$ matrices:

$$\mathfrak{sl}_{2} = \mathsf{span}_{\mathbb{R}}\left\{ \sigma_{1},\sigma_{2},\sigma_{3}\right\}.$$

Again recall that the representation $\pi$ of $\mathfrak{gl}_{2}$ on $\mathbb{C}^{2}$ is just the identity map:

$$\mathsf{id} : \mathfrak{gl}_{2} \rightarrow \mathfrak{gl}_{2}.$$

Once again, as $\mathfrak{su}_{2}$ is a subalgebra of $\mathfrak{gl}_{2}$, the representation on $\mathbb{C}^{2}$ is just a restriction of $\pi$

$$\mathsf{id}\Big|_{\mathfrak{su}} : \mathfrak{gl}_{2} \rightarrow \mathfrak{gl}_{2},$$

or

$$\mathsf{id}: \mathfrak{su}_{2} \rightarrow \mathfrak{su}_{2}.$$

Of course, $\mathfrak{su}_{2}$ still acts on $\mathbb{C}^{2}$ through ordinary matrix multiplication.

The Lie group $\mathsf{SU}_{2}$

As a final example, consider the Lie group of special unitary matrices in two-dimensions: $\mathsf{SU}_{2}$. As you might recall, these are those invertible matrices of unit determinant whose adjoint is their inverse

$$\mathsf{SU}_{2} = \left\{ m \in \mathsf{GL}_{2}\mathbb{C}\;\Big|\; \det m = 1,\quad m^{\dagger} = m^{-1}\right\}.$$

By definition, $\mathsf{SU}_{2}$ is a subset of $\mathsf{SL}_{2}\mathbb{C}$. For $M$ and $N$ in $\mathsf{SU}_{2}$, the matrix (and group) product $MN$ is also unitary. To see this explicitly note that

$$(MN)^{\dagger} = (\overline{M}\overline{N})^{\mathsf T} = \overline{N}^{\sf T}\overline{M}^{\sf T} = N^{\dagger} M^{\dagger},$$

so that

$$(MN)^{\dagger} MN = N^{-1}M^{-1}MN = \mathbb{1},$$

so that

$$(MN)^{\dagger} = (MN)^{-1}.$$

Thus, $\mathsf{SU}_{2}$ is a subgroup of $\mathsf{GL}_{2}\mathbb{C}$, the automorphism group of $\mathbb{C}^{2}$.

By previous discussion, $\mathbb{C}^{2}$ can also serve as an $\mathsf{SU}_{2}$ module via the identity map, $\mathsf{id}$, so $\mathsf{SU}_{2}$ still acts on $\mathbb{C}^{2}$ through ordinary matrix multiplication.

These matrices play a particularly important role in the study of matrix representations of Lie groups. Our first hint of what that role is comes in the exercises.

Exercises

Exercise 8.1 The exponential map

The exponential map, which takes a Lie algebra $\mathfrak{g}$ to its Lie group $G$:

$$\exp : \mathfrak{g} \rightarrow G,$$

is defined explicitly via the formal sum

$$\exp: g \mapsto \sum_{n=0}^{\infty} \frac{1}{n!}g^{n}.$$

Does this sum always converge? How would you assess convergence, in this context? Use this construction to argue how a concrete $\mathfrak{g}$-module like $\mathbb{C}^{2}$ can be identified with the $G$-module.

Exercise 8.2 : $\mathfrak{gl}_{2}$ and $\mathfrak{sl}_{2}$

It is clear that $\mathfrak{sl}_{2}$ is a linear subspace of $\mathfrak{gl}_{2}$. What is the one (complex) dimensional subspace of $\mathfrak{gl}_{2}$ that is orthogonal to $\mathfrak{sl}_{2}$? Argue that $\mathfrak{sl}_{2}$ is a Lie subalgebra of $\mathfrak{gl}_{2}$. In terms of real vector spaces, what are those matrices in $\mathfrak{gl}_{2}$ orthogonal to the real subspace $\mathfrak{su}_{2}$?

Exercise 8.3 The one-dimensional Modules

Let $\mathbb{A}$ be any of $\mathbb{R},\mathbb{C}$ or $\mathbb{H}$. Argue that each of $\mathbb{A}$ is a Lie algebra and with the help of the previous lecture, construct the adjoint representation. Determine the automorphism group for each $\mathbb{A}$, and write down the associated Lie group for which $\mathbb{A}$ serves as a most general representation. Argue that all of $\mathbb{A}$ can serve as a module for both $\mathsf{End}\,\mathbb{R}$ and $\mathsf{Aut}\,\mathbb{R}$. Can you make $\mathbb{H}$ into a $\mathsf{End}\,\mathbb{C}$ or $\mathsf{Aut}\,\mathbb{C}$ module?

Exercise 8.4 The Lie group $\mathsf{U}_{1}$

Using the previous exercise, show that $\mathbb{C}$ is a module for $\mathsf{U}_{1}$. Argue that $\mathbb{C}^{n}$ for any $n$ is also a $\mathsf{U}_{1}$-module by directly constructing a representation for it. Is that representation unique?

Exercise 8.5 Unitary Representations

A matrix representation is said to be unitary if the inner product on the associated module $V$ is positive definite. That is, for any nonzero $v$ in $V$, the inner product with itself, $(v,v)$ is real and greater than zero. Given the action of $\pi(a)$, this implies that in particular,

$$(\pi(a)\cdot v , \pi(a)\cdot v) > 0.$$

For each of the group representations considered in this lecture, determine which are unitary. What is the most general, unitary representation on $\mathbb{C}^{n}$?

Exercise 8.6 Unitary Representations for $\mathsf{GL}_{2}\mathbb{R}$

Repeat our analysis of representations for $\mathbb{R}^{2}$. What is the most general, unitary representation on $\mathbb{R}^{2}$?

$^{1}$: These definitions are clear for concrete vector spaces such as $\mathbb{A}^{n}$. For an abstract vector space - perhaps something like smooth functions over the circle - a a similar notation is often used, although they need not be explicit matrices.

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